Next Greater Element I【第一个数组中数字在第二个数组中找次大值】

最新推荐文章于 2021-10-26 01:11:44 发布

Sim0Hayha

最新推荐文章于 2021-10-26 01:11:44 发布

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分类专栏：数据结构与算法

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本文介绍了一种高效算法，用于解决寻找数组中下一个更大的元素问题。通过使用栈来跟踪元素，该算法能够找到给定子数组中每个元素在主数组中的下一个更大值。文章详细解释了算法的工作原理及实现过程。

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PROBLEM:

You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1's elements in the corresponding places of nums2.

The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
Output: [-1,3,-1]
Explanation:
    For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
    For number 1 in the first array, the next greater number for it in the second array is 3.
    For number 2 in the first array, there is no next greater number for it in the second array, so output -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4].
Output: [3,-1]
Explanation:
    For number 2 in the first array, the next greater number for it in the second array is 3.
    For number 4 in the first array, there is no next greater number for it in the second array, so output -1.

Note:

All elements in nums1 and nums2 are unique.
The length of both nums1 and nums2 would not exceed 1000.

SOLVE:

class Solution:
    def nextGreaterElement(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        stack=[]
        dict_nums2={}
        for num in nums2:
            while len(stack) and stack[-1]<num:
                dict_nums2[stack[-1]]=num
                del stack[-1]
            stack.append(num)
        res=[]
        for num in nums1:
            res.append(-1 if num not in dict_nums2 else dict_nums2[num])    #python3已经放弃了has_key方法
        return res

思路：用一个堆栈保存nums2中的降序数列，然后按下标顺序遍历剩下的数字n，把比它小的数字pop出来，并且保存它和n的键值对，最后遍历nums1，如果元素在字典中存在，则为键对应值否则为-1.